Hello. I am stuck on a calculus problem and hope someone can help me. I was given a function for the value of a stock [ V(t) = Ke^(t^1/2), where K>0] and alternative investments [ A(t) = e^rt ]. I had to calculate the instantaneous rate of change for each, which I did, and also calculate the optimal time (t) to sell the stock. I cannot figure out this last part. How to I calculate the optimal time to sell the stock? Is it when V'(t)/V(t) = r? I guess I'm confused on how to relate the two results to find this? I tried setting the instantaneous rates of change equat to each other to find t, but the result didn't make sense: (1/(2(t^12)) = r. I tried to plug in an interest rate and got something weird. Can anyone help me?
2007-01-18 00:19:05 · 7 answers · asked by Myan 1 in Science & Mathematics ➔ Mathematics
Instantaneous Rate of Change can be tricky. In economics, we often are looking for a percentage rather than just the rate of change (first derivative), such as the change in income at time t (this would be I'(t)/I(t)). You look like to have the right idea. Take the first derivative of each, like you said, and divide each by the function itself (I think you indicated this above). Next, set the results (which are the instantaneous rates of change) equal to each other. We know off the bat that A'(t) = r. You want to sell the stock when the instantaneous rate of change of the stock is equal to the rate earned on the alternate investments (r). At that moment, t, the value of the stock and the value of the alternate investments are increasing at the same rate. I hope that makes sense.
2007-01-19 06:43:31 · answer #1 · answered by ? 3 · 0⤊ 0⤋
I do not understand this problem much but I think that the best time to sell the stocks has nothing to do with individual maximum or minimum of the two investments. The optimal time to sell the stocks and invest in the alternate place will be when the difference in the values of your stocks and alternative investment is maximum. That will let you get the most benefit from your alternative investment. So I suggest you try to maximize the function f(t)=V(t)-A(t). Don't forget to check the second derivative to ensure maximum. Hope that gives some direction...
2007-01-18 01:05:19 · answer #2 · answered by Defunct 2 · 0⤊ 0⤋
If r is positive also, both investments are increasing in value all the time. You would want to shift from the V investment to the A investment when the A investment is rising faster. You were correct in your procedure to set the first derivatives equal to each other and solve for t. You claim to have arrived at
(1/(2(t^12)) = r
but that can't be right. V' contains K as a factor and A' contains r as a factor. You would solve for t in terms of K and r, by using logs.
A'(t) = re^rt
V'(t) = Ke^(t^1/2) * (1/2)t^(-1/2)
NoSignal is wrong to think you would wait for investment V to reach a maximum. There is no maximum. V rises indefinitely. You correctly want to shift when investment A starts to rise faster than V.
Defunct is wrong to think you want to maximize the difference between the values of the investments. Why should you care about the difference? It's the rate of increase that is important.
2007-01-18 00:52:59 · answer #3 · answered by ? 6 · 0⤊ 0⤋
You need to use the derivative to find the maximim of your function(s). The maximum value is where you will get the highest amount for the investment.
2007-01-18 00:44:49 · answer #4 · answered by Anonymous · 0⤊ 1⤋
find minimum of t using first and second derivatives of the given function
2007-01-18 00:55:47 · answer #5 · answered by iyiogrenci 6 · 0⤊ 0⤋
f'(x)=(2x(x+1)-(x^2-2))/(x+1)^2 f'(3)=(2*3(3+1)-(3^2-2))/(3+1)^2 =(24-7)/16 =17/16
2016-05-24 03:02:52 · answer #6 · answered by Alejandra 4 · 0⤊ 0⤋
Sorry wish I could help.
2007-01-18 00:40:07 · answer #7 · answered by The Story Teller 2 · 0⤊ 1⤋